2011 AMC 8 Problems/Problem 22

Problem

What is the tens digit of $7^{2011}$ ?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Solution 1

The first couple powers of $7$ are $7, 49, 343, 2401, 16807.$ As you can see, the last two digits cycle after every 4 powers. $7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).$ From there, we go two more powers. The last two digits are $43$ so the tens digit is $\boxed{\textbf{(D)}\ 4}$

Solution 2

We want the tens digit So, we take $7^{2011}\ (\text{mod }100)$ . That is congruent to $7^{11}\ (\text{mod}100)$ . From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 4

2011 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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2011 AMC 8 Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

See Also